Differential Equations Test

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Differential Equations Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following differential equations has y = x as one of its particular solution?

A
$$\cfrac{d^2y}{dx^2}-x^2\cfrac{dy}{dx}+xy=x$$

B
$$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=x$$

C
$$\cfrac{d^2y}{dx^2} -x^2\cfrac{dy}{dx}+xy=0$$

D
$$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=0$$

Solution

$$y=x\\
\dfrac{dy}{dx}=1\\
\dfrac{d^2y}{dx^2}=0$$
Now putting these values in given options then we will find that
Option C is correct
$$\dfrac{d^2y}{dx^2}-x^2\dfrac{dy}{dx}+xy=0$$
Question 2
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The equation of a curve passing through $$\displaystyle \left( 2,\frac { 7 }{ 2 } \right) $$ and having gradient $$\displaystyle 1-\frac { 1 }{ { x }^{ 2 } } $$ at $$(x,y)$$ is:

A
$$y={ x }^{ 2 }+x+1$$

B
$$xy={ x }^{ 2 }+x+1$$

C
$$xy=x+1$$

D
None of these

Solution

We have, $$\displaystyle \frac { dy }{ dx } =1-\frac { 1 }{ { x }^{ 2 } } \Rightarrow y=x+\frac { 1 }{ x } +c$$
This passes through $$\displaystyle \left( 2,\frac { 7 }{ 2 } \right) $$, therefore
$$\displaystyle \frac { 7 }{ 2 } =2+\frac { 1 }{ 2 } +c\Rightarrow c=1$$
Thus the equation of the curve is
$$\displaystyle y=x+\frac { 1 }{ x } +1\Rightarrow xy={ x }^{ 2 }+x+1$$
Question 3
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Which of the following differential equations has $$y = c_1 e^x + c_2 e^{x}$$ as the general solution?

A
$$\dfrac{d^2y}{dx^2}+y = 0$$

B
$$\dfrac{d^2y}{dx^2}-y = 0$$

C
$$\dfrac{d^2y}{dx^2}+1 = 0$$

D
$$\dfrac{d^2y}{dx^2}-1 = 0$$

Solution

$$y=c_1e^x+c_2e^x\\
\dfrac{dy}{dx}=c_1e^x+c_2e^x\\
\dfrac{d^2y}{dx^2}=c_1e^x+c_2e^x\\
\therefore \dfrac{d^2y}{dx^2}-y=0$$
Question 4
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The solution to the differential equation $$(x + 1) \displaystyle \frac{dy}{dx} - y = e^{3x} (x + 1)^2 $$ is

A
$$y = (x + 1)e^{3x} + c$$

B
$$3y = (x + 1) + e^{3x} + c$$

C
$$\displaystyle \frac{3y}{x+ 1} = e^{3x} + c$$

D
$$ye^{-3x} = 3 (x + 1) + c$$

Solution

The given equation is $$\displaystyle \dfrac{dy}{dx} - \dfrac{y}{x + 1} = e^{3x} (x+ 1)$$
I.F. $$= \displaystyle e^{\displaystyle \int - \dfrac{1}{x + 1} dx} = e^{\displaystyle - log (x + 1)} = \dfrac{1}{x + 1}$$
The solution is
$$\displaystyle y \left ( \dfrac{1}{x +1} \right ) = \int e^{3x} (x +1). \dfrac{1}{x + 1} dx + a$$
$$\Rightarrow \displaystyle \dfrac{y}{x + 1} = \int e^{3x} dx+ a = \dfrac{e^{3x}}{3} + a$$
$$\Rightarrow \displaystyle \dfrac{3y}{x + 1} = e^{3x} + c, c = 3a$$
Question 5
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The solution of differential equation
$$\displaystyle \frac { dy }{ dx } =-\left( \frac { y+\sin { x } }{ x } \right) $$ satisfying condition $$\displaystyle y(0) = 1$$, is

A
$$\displaystyle \cos x = xy - 1$$

B
$$\displaystyle \cos x = xy + 1$$

C
$$\displaystyle \cos x = xy $$

D
$$\displaystyle \cos x = x + 1$$

Solution

$$\displaystyle xdy=-ydx-\sin { x } dx$$
$$\displaystyle \Rightarrow -\sin { x } dx=xdy+ydx$$
$$\displaystyle \Rightarrow -\sin { x } dx=d\left( xy \right) $$
$$\displaystyle \Rightarrow \int { -\sin x { dx } } =\int { d\left( xy \right) } $$
$$\displaystyle \Rightarrow \cos { x } =xy+c$$
$$\displaystyle y(0)=1\Rightarrow c=1$$
$$\displaystyle \Rightarrow \cos { x } =xy+1$$
Question 6
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The solution of the equation $$\displaystyle \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ x }+{ e }^{ -x }$$ is-
Note : (where c & d are arbitrary constants in the given options)

A
$$\displaystyle y={ e }^{ x }-{ e }^{ -x }+cx+d$$

B
$$\displaystyle y={ e }^{ x }+{ e }^{ -x }+cx+d$$

C
$$\displaystyle y=-{ e }^{ x }+{ e }^{ -x }+cx+d$$

D
None of these

Solution

Given, $$\displaystyle \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ e }^{ x }+{ e }^{ -x }$$
Integrating gives
$$\Rightarrow \displaystyle \frac { dy }{ { dx } } ={ e }^{ x }-{ e }^{ -x }-c$$
Again integrating,
$$\displaystyle \Rightarrow y={ e }^{ x }+{ e }^{ -x }-cx+d$$
Question 7
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Solution of the differential equation
$$\displaystyle \left( 2x-y+2 \right) dx+\left( 4x-2y-1 \right) dy=0$$ is

A
$$\displaystyle 2x-y={ ce }^{ -\left( x+2y \right) }$$

B
$$\displaystyle 2x+y={ ce }^{ \left( 2x+y \right) }$$

C
$$\displaystyle x-2y={ ce }^{ -\left( x+2y \right) }$$

D
$$\displaystyle 2x+y={ ce }^{ \left( x+2y \right) }$$

Solution

Given equation is $$\displaystyle \frac { dy }{ dx } =\frac { -\left\{

\left( 2x-y \right) +2 \right\} }{ 2\left( 2x-y \right) -1 } $$
Put $$\displaystyle 2x-y=v\Rightarrow \displaystyle 2-\frac { dy }{ dx } =\frac { dv }{ dx } $$

$$\Rightarrow\displaystyle -\left( \frac { dv }{ dx } -2 \right) =-\left( \frac { v+2 }{ 2v-1 } \right) $$

$$\displaystyle

\Rightarrow \frac { dv }{ dx } =\left( \frac { v+2 }{ 2v-1 } +2 \right)

\Rightarrow \frac { dv }{ dx } =\frac { 5v }{ 2v-1 } $$

$$\displaystyle \Rightarrow \int { \frac { 2v-1 }{ 5v } } dv=\int { dx } $$

$$\displaystyle \Rightarrow \frac { 2 }{ 5 } v-\frac { 1 }{ 5 } \log { v } =x+\log { c } $$

$$\displaystyle

\Rightarrow \frac { 2 }{ 5 } \left( 2x-y \right) -\frac { 1 }{ 5 } \log

{ \left( 2x-y \right) } =x+\log { c } $$

$$\displaystyle \Rightarrow 2\left( 2x-y \right) -\log { \left( 2x-y \right) } =5x+5\log { c } $$

$$\displaystyle \Rightarrow 4x-2y-\log { \left( 2x-y \right) } =5x-\log { c } $$

$$\displaystyle \Rightarrow \log { c } -\log { \left( 2x-y \right) } =\left( x+2y \right) $$

$$\displaystyle\Rightarrow \log { \left( \frac { c }{ 2x-y } \right) } =\left( x+2y \right) $$

$$\displaystyle \Rightarrow\left( \frac { c }{ 2x-y } \right) ={ e }^{ \left( x+2y \right) }$$

$$\displaystyle\Rightarrow c.{ e }^{ -\left( x+2y \right) }=\left( 2x-y \right) $$
Question 8
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Solution of differential equation $$\displaystyle \frac { dx }{ dy } =\tan { x } \left( 1+ysinx \right) $$ is given by -

A
$$\displaystyle cosec x= -y+1+{ Ce }^{ -y }$$

B
$$\displaystyle y=\tan { x } +{ Ce }^{ X }$$

C
$$\displaystyle \sin { x } { e }^{ y }=1+y+C$$

D
$$\displaystyle cosecx=y+{ Ce }^{ y }$$

Solution

$$\displaystyle \cot { x }* \csc x\frac { dx }{ dy } = \csc x + y
$$
Put $$ \csc x=t$$
$$\displaystyle \Rightarrow -\csc x \ * \cot x\frac { dx }{ dy } =\frac { dt }{ dy } $$
Now D.E is $$\displaystyle \frac { dt }{ dy } =-t-y$$

I.F. =$$\displaystyle { e }^{\int dy }{ ={ e }^{ y } } $$
solution is $$\displaystyle t.{ e }^{ y }=\int { -y{ e }^{ y }dy } =-ye^y+e^y+c $$
$$\displaystyle \Rightarrow \csc x=-y+1+Ce^{-y}$$
Question 9
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The general solution of the differential equation $$\dfrac{y\, dx-x\, dy}{y}=0$$ is:

A
$$xy = C$$

B
$$x = Cy^2$$

C
$$y=Cx$$

D
$$y=Cx^2$$

Solution

$$\dfrac{ydx-xdy}{y}=0\\
dx-\dfrac{x}{y}dy=0\\
dx=\dfrac{x}{y}dy\\
\dfrac{dx}{x}=\dfrac{dy}{y}\\
lnx=lny+lnc\\
x=yc$$
Question 10
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The solution of the differential equation $$\dfrac { dy }{ dx } =\dfrac { x-y+3 }{ 2\left( x-y \right) +5 } $$ is

A
$$2\left( x-y \right) +\log { \left( x-y \right) } =x+c$$

B
$$2\left( x-y \right) -\log { \left( x-y+2 \right) } =x+c$$

C
$$2\left( x-y \right) +\log { \left( x-y+2 \right) } =x+c$$

D
None of the above

Solution

Given differential equation is
$$\dfrac { dy }{ dx } =\dfrac { x-y+3 }{ 2\left( x-y \right) +5 } $$
Let $$x-y=v\Rightarrow \dfrac { dy }{ dx } =1-\dfrac { dv }{ dx } $$
$$\therefore \dfrac { dv }{ dx } =\dfrac { v+2 }{ 2v+5 } $$
$$\Rightarrow \displaystyle\int { \dfrac { 2v+5 }{ v+2 } dv } =\displaystyle\int { dx } $$
$$\Rightarrow \displaystyle\int { \left( 2+\dfrac { 1 }{ v+2 } \right) dv } =\displaystyle\int { dx } $$
$$\Rightarrow 2v+\log { \left( v+2 \right) } =x+c$$
$$\Rightarrow 2\left( x-y \right) +\log { \left( x-y+2 \right) } =x+c$$

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Differential Equations Test - 28

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Question 1

$$\cfrac{d^2y}{dx^2}-x^2\cfrac{dy}{dx}+xy=x$$

$$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=x$$

$$\cfrac{d^2y}{dx^2} -x^2\cfrac{dy}{dx}+xy=0$$

$$\cfrac{d^2y}{dx^2}+x\cfrac{dy}{dx}+xy=0$$

Solution

Question 2

$$y={ x }^{ 2 }+x+1$$

$$xy={ x }^{ 2 }+x+1$$

$$xy=x+1$$

None of these

Solution

Question 3

$$\dfrac{d^2y}{dx^2}+y = 0$$

$$\dfrac{d^2y}{dx^2}-y = 0$$

$$\dfrac{d^2y}{dx^2}+1 = 0$$

$$\dfrac{d^2y}{dx^2}-1 = 0$$

Solution

Question 4

$$y = (x + 1)e^{3x} + c$$

$$3y = (x + 1) + e^{3x} + c$$

$$\displaystyle \frac{3y}{x+ 1} = e^{3x} + c$$

$$ye^{-3x} = 3 (x + 1) + c$$

Solution

Question 5

$$\displaystyle \cos x = xy - 1$$

$$\displaystyle \cos x = xy + 1$$

$$\displaystyle \cos x = xy $$

$$\displaystyle \cos x = x + 1$$

Solution

Question 6

$$\displaystyle y={ e }^{ x }-{ e }^{ -x }+cx+d$$

$$\displaystyle y={ e }^{ x }+{ e }^{ -x }+cx+d$$

$$\displaystyle y=-{ e }^{ x }+{ e }^{ -x }+cx+d$$

None of these

Solution

Question 7

$$\displaystyle 2x-y={ ce }^{ -\left( x+2y \right) }$$

$$\displaystyle 2x+y={ ce }^{ \left( 2x+y \right) }$$

$$\displaystyle x-2y={ ce }^{ -\left( x+2y \right) }$$

$$\displaystyle 2x+y={ ce }^{ \left( x+2y \right) }$$

Solution

Question 8

$$\displaystyle cosec x= -y+1+{ Ce }^{ -y }$$

$$\displaystyle y=\tan { x } +{ Ce }^{ X }$$

$$\displaystyle \sin { x } { e }^{ y }=1+y+C$$

$$\displaystyle cosecx=y+{ Ce }^{ y }$$

Solution

Question 9

$$xy = C$$

$$x = Cy^2$$

$$y=Cx$$

$$y=Cx^2$$

Solution

Question 10

$$2\left( x-y \right) +\log { \left( x-y \right) } =x+c$$

$$2\left( x-y \right) -\log { \left( x-y+2 \right) } =x+c$$

$$2\left( x-y \right) +\log { \left( x-y+2 \right) } =x+c$$

None of the above

Solution

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